Question

If $PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ such that $OPQ$ is an equilateral triangle, $O$ being the centre of the hyperbola, then the eccentricitye of the hyperbola satisfies

• A. $1<e<\frac{2}{\sqrt{3}}$
• B. $e=\frac{2}{\sqrt{3}}$
• C. $e=\frac{\sqrt{3}}{2}$
• D. $e>\frac{2}{\sqrt{3}}$

Answer 1

Answer: (D)

Let the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and any double ordinate $PQ$ be ( $a\sec \theta ,b\tan \theta )$, ( $\sec \theta ,-b$ $\tan \theta )\cdot O$ is the centre $(0,0)$.
$△OPQ$ is equilateral.

So $\tan 30^{\circ }=\frac{b\tan \theta }{a\sec \theta }$
$\Rightarrow 3\frac{b^2}{a^2}={\mathrm{cosec}}^2\theta$
$\Rightarrow 3\left(e^2-1\right)={\mathrm{cosec}}^2\theta$
$\text{ Now }{\mathrm{cosec}}^2\theta \ge 1$
Now ${\mathrm{cosec}}^2\theta \ge 1$
$\Rightarrow 3\left(e^2-1\right)\ge 1\Rightarrow e^2\ge \frac{4}{3}\Rightarrow e>\frac{2}{\sqrt{3}}$